Kienastsrx

2022-03-16

What is the instantaneous rate of change of $f\left(x\right)=\frac{x}{-x+7}$ at x=0?

nida0694ii5

Beginner2022-03-17Added 6 answers

The instantaneous rate of change is the first derivative calculated at

x=0

Hence

$f}^{\prime}\left(x\right)=\frac{7}{{(7-x)}^{2}$

so$f}^{\prime}\left(0\right)=\frac{7}{{7}^{2}}=\frac{1}{7$

x=0

Hence

so

Talaminiu2d

Beginner2022-03-18Added 4 answers

Apply the quotient rule, which is:

$\frac{d}{dx}\frac{f\left(x\right)}{g\left(x\right)}=\frac{-f\left(x\right)\frac{d}{dx}g\left(x\right)+g\left(x\right)\frac{d}{dx}f\left(x\right)}{{g}^{2}\left(x\right)}$

f(x)=x and g(x)=7-x

To find$\frac{d}{dx}f\left(x\right):$

Apply the power rule: x goes to 1

To find$\frac{d}{dx}g\left(x\right):$

Differentiate 7-x term by term:

The derivative of the constant 7 is zero.

The derivative of a constant times a function is the constant times the derivative of the function.

Apply the power rule: x goes to 1

So, the result is: -1

The result is: -1

Now plug in to the quotient rule:

$\frac{7}{{(7-x)}^{2}}$

Now simplify:

$\frac{7}{{(x-7)}^{2}}$

If we putting the value of x, we get:

0

f(x)=x and g(x)=7-x

To find

Apply the power rule: x goes to 1

To find

Differentiate 7-x term by term:

The derivative of the constant 7 is zero.

The derivative of a constant times a function is the constant times the derivative of the function.

Apply the power rule: x goes to 1

So, the result is: -1

The result is: -1

Now plug in to the quotient rule:

Now simplify:

If we putting the value of x, we get:

0

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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